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This is an attempt to derive the Taylor expansion at $0$ of the Lambert $w$ function (the branch containing the origin of the solution $w$ to the functional equation $w(x)e^{w(x)}=x$) without resorting to Lagrange inversion formula, nor complex analysis.

Let the superscript $^{(n)}$, (superscript $^{'}$) denote the $n$-th derivative with respect to $x$ (to $w$ respectively) . It is easy to see that \begin{align*} w^{(1)}=\frac{w}{x}\frac{1}{w+1} .\end{align*} Then by iterating the derivation process we obtain \begin{align*} &w^{(2)}=-\frac{w^2}{x^2}\frac{w+2}{(w+1)^3} \\ &w^{(3)}=\frac{w^3}{x^3}\frac{2w^2+8w+9}{(w+1)^5}\\ &\qquad\qquad..... \end{align*} Now we start an induction process and we suppose that there is some polynomial $P_n$ such that \begin{align*} &w^{(n)}=-\frac{(-w)^n}{x^n(w+1)^{2n-1}}P_n(w) \end{align*} After some algebra we see that these polynomials must obey the following recursion:

\begin{align*} P_{1}(w)&=1\\P_{n+1}(w)&=\big(n(w +3)-1\big)P_n(w)-(w+1)P_n^{'}(w). \end{align*}

What is the name of these polynomials?

How can it be shown that $P_n(0)=n^{n-1}$?

First instances: \begin{align*} P_1(w)&= \color{red}{1}\\ P_2(w)&=w+\color{red}{2}\\ P_3(w)&=2w^2+8w+\color{red}{9}\\ P_4(w)&=6w^3+36w^2+79w+\color{red}{64}\\ P_5(w)&=24w^4+192w^3+622w^2+974w+\color{red}{625}\\ \end{align*}

Or consider the triangle of numbers $a_{n,k}$ obtained recursively, such that the entries in line $n$ are computed with three consecutive entries from line $n-1$ by $$ a_{n,k}=n a_{n-1,k-1}+(3n-(k+1))a_{n-1,k}-(k+1)a_{n-1,k+1}$$ with initial conditions $ a_{n,-1}= 0$ , $ a_{0,0}= 1$ and $ a_{0,k}= 0$ for $k>0$.

This makes it clear that $a_{n,n}=n!$, but how could we obtained the closed form $a_{n,0}=(n+1)^n$ ?

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  • $\begingroup$ oeis.org/A042977 "Triangle $T(n,k)$ read by rows: coefficients of a polynomial sequence occurring when calculating the n-th derivative of Lambert function $W$." $\endgroup$ – Travis May 6 at 22:28
  • $\begingroup$ These polynomials also appear as factors of the expressions in Example 4.3 of Kruchinin, V., "Derivation of Bell Polynomials of the Second Kind" arxiv.org/abs/1104.5065 $\endgroup$ – Travis May 6 at 22:30
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This answer does not fully address the question of how to compute an explicit formula from the recursion relation (though it indicates a reference which contains a starting point), and so is not a proper answer, but it is too long for a comment, and I expect it might still be useful.

It follows from a comment by Vladimir Kruchinin in OEIS A042977, "Triangle $T(n,k)$ read by rows: coefficients of a polynomial sequence occurring when calculating the $n$th derivative of Lambert function $W$" that $$\color{#df0000}{\boxed{P_n(w) = \sum_{m = 0}^n \left[ \sum_{j=0}^m {2 n + 1 \choose m - j} \sum_{k=0}^j (-1)^k \frac{(n+k+1)^{n+j}}{(j-k)! k!} \right] w^m}} .$$ See under the heading $\texttt{FORMULA}$, and NB that the expression in this answer differs from the one in OEIS entry by a factor of $(-1)^n$ to account for the formulation in the question. This expression can be deduced from Example 4.3 in Kruchinin's preprint "Derivation of Bell Polynomials of the Second Kind".

In particular, to compute the constant term of $P_n(w)$, we can evaluate the inner sum for $j = 0$. Both summations have only a single term, and it simplifies to the conjectured formula, $$\color{#df0000}{\boxed{P_n(0) = (n + 1)^n}} .$$

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  • $\begingroup$ This is very relevant and useful, and since I requested for reference, I accept it as an answer; but I agree that it is not entirely satisfactory: the formula for $P_n(w)$ was obtained by making use of the Lambert w function. Given the recursion only, without any other contextual info, would there be a proof that does not make use of Lambert w ? $\endgroup$ – René Gy May 7 at 22:55
  • $\begingroup$ I'm glad you found the answer useful! I don't know any general methods for solving recursion formulas of the form of yours. You might even formulate a new question asking just about methods for solving the recursion. (If you do this, I suggest you link to this question for context.) $\endgroup$ – Travis May 7 at 22:58

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